Sample Problems. Lecture Notes Equations with Parameters page 1

Transcription

1 Lecture Notes Equations with Parameters page Sample Problems. In each of the parametric equations given, nd the value of the parameter m so that the equation has exactly one real solution. a) x + mx m 0 b) mx + mx + m x + c) x + m x. Consider the parametric equation 5m 3x 5mx + 6m + x 0. a) Find all values of m for which x 7 is a solution of the equation. b) Find all values of m for which there are two di erent real solutions of this equation. c) Find all values of the parameter m for which the two solutions of the equation add up to 3. d) Find all values of the parameter m for which the product of the two solutions of the equation is Consider the equation x + mx + 3mx + 5x. a) Find all values of m for which the equation has exactly one real solution. b) Find all values of m for which the equation has no real solution. c) Find all values of m for which the two real solutions x and x are such that x + x :. Consider the parametric equation m + mx + mx x a) Find all values of m for which x is a solution of the equation. b) Find all values of m for which there are two real solutions of the equation. c) Find all values of m for which there are two real solutions, x and x of the equation such that x + x 5. c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009

4 Lecture Notes Equations with Parameters page c) x + m x 0; Solution : Completing the square. First we separately consider the case for which the equation is linear, namely, m 0. If m 0, our equation is x which indeed has exactly one solution. Now, if m 6 0; we multiply both sides by x and complete the square x + m x multiply by x x + m x subtract x x x + m 0 factor out x x + m 0 x x x + 6! x x + {z 6} 6 + m 0! x 6 + m 0 To have exactly one solution, the part after the complete square must be equal to zero. 6 + m 0 multiply by 6 + m 0 m m There is one more possibility we need to consider. The original equation is x+ m x and NOT x x+m 0. We might get exactly one solution as follows: the discriminant is positive, indicating two solutions, but one of the solutions is 0 which would have to be ruled out as it is not in the domain of the original equation. To check out this case, we take x x + m 0 and see what m is so that one of the solutions for x is zero. x x + m 0 (0) 0 + m 0 m 0 We have already considered this case, and so our solution, m 0, m is complete. Solution : Quadratic Formula First we separately consider the case for which the equation is linear, namely, m 0. If m 0, our equation is x which indeed has exactly one solution. Now, if m 6 0; we multiply both sides by x and use the formula: x + m x multiply by x x + m x subtract x x x + m 0 Now a, b, and c m: To have exactly one solution, the discriminant needs to be zero. We solve this equation for m. b ac 0 m ( ) () m 0 m m 0 c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009

5 Lecture Notes Equations with Parameters page 5 There is one more possibility we need to consider. The original equation is x + m and NOT x x x + m 0. We might get exactly one solution as follows: the discriminant is positive, indicating two solutions, but one of the solutions is 0 which would have to be ruled out as it is not in the domain of the original equation. To check out this case, we take x x + m 0 and see what m is so that one of the solutions for x is zero. x x + m 0 (0) 0 + m 0 m 0 We have already considered this case, and so our solution, m 0, m is complete.. Consider the parametric equation 5m 3x 5mx + 6m + x 0. a) Find all values of m for which x 7 is a solution of the equation. ; Solution: Just plug in x 7 into the equation and see what that gives us for m. 5m 3x 5mx + 6m + x 0 5m 3 (7) 5m (7) + 6m + (7) 0 5m 35m + 6m m 30m + 0 factor out 6 6 m 5m (m ) (m ) 0 m m b) Find all values of m for which there are two di erent solutions of this equation. m 6 5 Solution: We rearrange the polynomial by degrees of the variable x. x + x ( 5m 3) + 6m + 5m 0. Now a, b 5m 3, and c 6m + 5m. For two solutions, we need the discriminant, b ac to be positive. b ac > 0 ( 5m 3) () 6m + 5m > 0 5m + 30m + 9 m 0m + 6 > 0 m + 0m + 5 > 0 (m + 5) > 0 m 6 5 c) Find all values of the parameter m for which the two solutions of the equation add up to 3. Solution: the sum of the two solutions is b a. b a 3 ( 5m 3) 3 5m m 0 m c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009

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